Physics 3 Marks Question

$\text{C}_1=8\mu\text{F},\ \text{C}_2=4\mu\text{F},\ \text{C}_3=4\mu\text{F}$

$\text{C}_{\text{eq}}=\frac{(\text{C}_2+\text{C}_3)\times\text{C}_1}{\text{C}_1+\text{C}_2+\text{C}_3}$

$=\frac{8\times8}{16}=4\mu\text{F}$

Since B & C are parallel & are in series with A.

So, $\text{q}_1=8\times6=48\mu\text{C}$

$\text{q}_2=4\times6=24\mu\text{C}$

$\text{q}_3=4\times6=24\text{C}\mu$

$\therefore\text{C}_1=4\mu\text{F},\ \text{C}_2=6\mu\text{F},\ \text{V}=20\text{V}$

Eq. capacitor $\text{C}_{\text{eq}}=\frac{\text{C}_1\text{C}_2}{\text{C}_1+\text{C}_2}=\frac{4\times6}{4+6}=2.4$

$\therefore$ The Eq Capacitance $\text{C}_{\text{eq}}=2.5\mu\text{F}$

$\therefore$ The energy supplied by the battery to each plate

$\text{E}=\Big(\frac{1}{2}\Big)\text{CV}^2$

$=\Big(\frac{1}{2}\Big)\times2.4\times20^2=480\mu\text{J}$

$\therefore$ The energy supplies by the battery to capacitor $=2\times480=960\mu\text{J}$

$\text{C}_1=2\mu\text{F},\ \text{C}_2=4\mu\text{F},\ \text{C}_3=6\mu\text{F}$

$\text{V}=12\text{V}$

$\text{cq}=\text{C}_1+\text{C}_2+\text{C}_3$

$=2+4+6=12\mu\text{F}$

$=12\times10^{-6}\text{F}$

$\text{q}_1=12\times2=24\mu\text{C}$

$\text{q}_2=12\times4=48\mu\text{C}$

$\text{q}_3=12\times6=72\mu\text{C}$

Here three capacitors are formed

And each of $\text{A}=\frac{96}{\in_0}\times10^{-12}\text{f.m.}$

$\text{d}=4\text{mm}=4\times10^{-3}\text{m}$

$\therefore$ Capacitance of a capacitor

$\text{C}=\frac{\in_0\text{A}}{\text{d}}=\frac{\in_0\frac{96\times10^{-12}}{\in_0}}{4\times10^{-3}}=24\times10^{-9}\text{F}$

$\therefore$ As three capacitor are arranged is series

So, $\text{C}_\text{eq}=\frac{\text{C}}{\text{q}}=\frac{24\times10^{-9}}{3}=8\times10^{-9}$

$\therefore$ The total charge to a capacitor = 8 × 10^-9 × 10 = 8 × 10^-8c

$\therefore$ The charge of a single Plate = 2 × 8 × 10^-8 = 16 × 10^-8 = 0.16 × 10^-6 = 0.16μc.

$\text{V}=\text{10v}$

$\text{C}_{\text{eq}}=\text{C}_1+\text{C}_2$ $[\therefore$ They are parallel$]$

$=5+6=11\mu\text{F}$

$\text{q}=\text{CV}=11\times10=110\mu\text{C}$

$\text{A}=25\text{cm}^2=2.5\times10^{-3}\text{cm}^2$

$\text{d}=1\text{mm}=0.01\text{m}$

$\text{V}=\text{6V},\ \text{Q}=1$

$\text{C}=\frac{\in_0\text{A}}{\text{d}}=\frac{8.854\times10^{-12}\times2.5\times10^{-3}}{0.01}$

$\text{Q}=\text{CV}=\frac{8.854\times10^{-12}\times2.5\times10^{-3}}{0.01}\times6$

$=1.32810\times10^{-10}\text{C}$

$\text{W}=\text{Q}\times\text{V}=1.32810\times10^{-10}\text{C}\times6$

$=8\times10^{-10}\text{J}.$

$\therefore$ Area $=\pi\text{R}^2$

$\text{C}=1\text{f}$

$\text{D}=1\text{mm}=10^{-3}\text{m}$

$\therefore\text{C}=\frac{\in_0\text{A}}{\text{d}}$

$\Rightarrow1=\frac{8.85\times10^{-12}\times\pi\text{r}^2}{10^{-3}}$

$\Rightarrow\text{r}^2=\frac{10^{-3}\times10^{12}}{8.85\times\pi}=\frac{10^9}{27.784}$

$=5998.5\text{m}=6\text{Km}$

$\therefore$ The net potential difference = 10L.

$\therefore$ Given that

Capacitance $=10\mu\text{F}$

Charge $=20\mu\text{C}$

$\therefore$ The effective charge $=\frac{20-0}{2}=10\mu\text{F}$

$\therefore\text{C}=\frac{\text{q}}{\text{V}}$

$\Rightarrow\text{V}=\frac{\text{q}}{\text{C}}=\frac{10}{10}=1\text{V}$